package SubjectTree.Three;

import java.util.Deque;
import java.util.LinkedList;
import java.util.Queue;

import Utility.TreeNode;

public class HasPathSum {

/**
 * 难度：简单
 * 
 * 112. 路径总和
 * 	给你二叉树的根节点 root 和一个表示目标和的整数 targetSum ，判断该树中是否存在 
 * 	根节点到叶子节点 的路径，这条路径上所有节点值相加等于目标和 targetSum 。
 * 	叶子节点 是指没有子节点的节点。
 * 	
 * 示例 1：
 * 	输入：root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
 * 	输出：true
 * 	
 * 示例 2：
 * 	输入：root = [1,2,3], targetSum = 5
 * 	输出：false
 * 	
 * 示例 3：
 * 	输入：root = [1,2], targetSum = 0
 * 	输出：false
 * 	
 * 提示：
 * 	树中节点的数目在范围 [0, 5000] 内
 * 	-1000 <= Node.val <= 1000
 * 	-1000 <= targetSum <= 1000
 *
 * */
	
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		HasPathSum hps = new HasPathSum();
		System.out.println(hps.hasPathSum_1(TreeNode.MkTree("[5,4,8,11,null,13,4,7,2,null,null,null,1]"), 22));
	}
	//自己写(逐层遍历)
	public boolean hasPathSum(TreeNode root, int targetSum) {
		if(root == null) return false;
		Deque<TreeNode> deque = new LinkedList<>();
		Queue<Integer> queue = new LinkedList<>();
		deque.offer(root);
		queue.offer(root.val);
		while(!deque.isEmpty()) {
			int currentLevelSize = deque.size();
			for(int i=0;i<currentLevelSize;i++) {
				TreeNode node = deque.poll();
				int count = queue.poll();
				if(node.left==null&&node.right==null) {
					if(count == targetSum)return true;
				}
				if(node.left!=null) {
					deque.offer(node.left);
					queue.offer(count+node.left.val);
				}
				if(node.right!=null) {
					deque.offer(node.right);
					queue.offer(count+node.right.val);
				}
			}
		}
		return false;
    }
	//自己写(递归)
	public boolean hasPathSum_1(TreeNode root, int targetSum) {
		return traversal(root, targetSum, 0);
    }
	public boolean traversal(TreeNode root, int targetSum, int count) {
		if(root == null) return false;
		boolean leftbool = false;
		boolean rightbool = false;
		count += root.val;
		if(root.left==null && root.right==null) {
			if(count == targetSum)return true;
		}
		if(root.left!=null) {
			leftbool = traversal(root.left, targetSum, count);
		}
		if(root.right!=null) {
			rightbool = traversal(root.right, targetSum, count);
		}
		return leftbool || rightbool;
	}
	//方法一：广度优先搜索
	public boolean hasPathSum1(TreeNode root, int sum) {
        if (root == null) {
            return false;
        }
        Queue<TreeNode> queNode = new LinkedList<TreeNode>();
        Queue<Integer> queVal = new LinkedList<Integer>();
        queNode.offer(root);
        queVal.offer(root.val);
        while (!queNode.isEmpty()) {
            TreeNode now = queNode.poll();
            int temp = queVal.poll();
            if (now.left == null && now.right == null) {
                if (temp == sum) {
                    return true;
                }
                continue;
            }
            if (now.left != null) {
                queNode.offer(now.left);
                queVal.offer(now.left.val + temp);
            }
            if (now.right != null) {
                queNode.offer(now.right);
                queVal.offer(now.right.val + temp);
            }
        }
        return false;
    }
	//方法二：递归
	public boolean hasPathSum2(TreeNode root, int sum) {
        if (root == null) {
            return false;
        }
        if (root.left == null && root.right == null) {
            return sum == root.val;
        }
        return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
    }
}
